综合模拟1
- 格式:doc
- 大小:303.50 KB
- 文档页数:5
一、选择题(每题3分,共48分)1.近代考古发现,陕西陇县边家庄秦墓、湖南常德楚墓及长沙楚墓等遗址,出土的铁器大都是春秋晚期的遗物,计有铁块、铁条、铁削、铁铧、铁铲、钢剑及铁鼎、铁带钩、铁环等铁制农具、日用生活品及武器。
这说明春秋晚期()A.冶铁技术有了明显进步B.民营冶铁业进一步发展C.粮食产量得到显著提高D.铁制农具已被普遍使用2.下面为先秦至汉盐价变化统计表,据此表可知()齐国、秦国汉文帝汉武帝东汉100钱/石120—150钱/石300—1 100钱/石400—8 000钱/石A.国家控制经济的力度增强B.秦汉时期的通货膨胀严重C.汉代的经济逐渐恢复发展D.秦汉人口有较大规模增加3.唐玄宗针对科举考试中的浮艳文风,下诏要求“自今以后,不得更然”;针对进士试中诗赋过于讲究声律的现象,下诏要求更加重视文词。
这些举措()A.彻底扭转了华丽空洞的文风B.改变了科举考试的主要科目C.更加有利于选拔高素质人才D.剥夺了宰相选官时的主导权4.公元995年,宋太宗下诏“改撰京城内外坊名八十余,由是分定布列”。
1002年,宋真宗下诏“复唐代长安禁鼓昏晓之制”。
到了宋神宗时期,汴京城已“不闻金鼓之声”,“坊市之名,多失标榜”。
材料反映了北宋()A.商品经济发展时空范围扩大B.政府强化了坊市制度C.城市安全管理有弱化的趋势D.加强了城市建设规划5.明代松江地区农家收支情况如表所示,对下表中的现象解读最合理的是()A.松江地区实现了均衡发展B.引进作物新品种提高了粮食产量C.传统经济结构发生了质变D.市镇经济的发展增加了农民收入6.1898年12月31日,京师大学堂开学,第一年课程只设诗、书、易、礼四堂和春、秋两堂。
第二年中西并学,除经史外,开设算学、格致、化学及英文、德文、法文、俄文、日文等普通课程,另立史学、地理、政治专门讲堂。
学堂的兴办()A.体现了维新派的改革愿望B.开创了近代教育的新体制C.反映了传统向近代的嬗变D.成为近代新式教育的开端7.下表所示为部分学者对近代中国民主革命中某一历史事件的评论。
专升本(高等数学一)综合模拟试卷1(题后含答案及解析)题型有:1. 选择题 2. 填空题 3. 解答题选择题1.极限等于( )A.eB.ebC.eabD.eab+b正确答案:C解析:由于,故选C。
知识模块:极限和连续2.在空间直角坐标系中,方程x2-4(y-1)2=0表示( )A.两个平面B.双曲柱面C.椭圆柱面D.圆柱面正确答案:A解析:由于所给曲面方程x2-4(y-1)2=0中不含z,可知所给曲面为柱面,但是由于所给方程可化为x2=4(y-1)2,进而可以化为x=2(y-1)与-z=2(y-1),即x-2y+2=0,x+2y-2=0,为两个平面,故选A。
知识模块:空间解析几何3.级数是( )A.绝对收敛B.条件收敛C.发散D.收敛性不能判定正确答案:A解析:依前述判定级数绝对收敛与条件收敛的一般原则,常常先判定的收敛性,由于的p级数,知其为收敛级数,因此所给级数绝对收敛,故选A。
知识模块:无穷级数填空题4.若函数在x=0处连续,则a=________。
正确答案:-2解析:由于(无穷小量乘有界变量),而f(0)=a+2,由于f(x)在x=0处连续,应有a+2=0,即a=-2。
知识模块:极限和连续5.若f’(x0)=1,f(x0)=0,则=________。
正确答案:-1解析:由于f’(x0)存在,且f(x0)=0,由导数的定义有知识模块:一元函数微分学6.设y=xe+ex+lnx+ee,则y’=________。
正确答案:y’=ee-1+ex+解析:由导数的基本公式及四则运算规则,有y’=ee-1+ex+。
知识模块:一元函数微分学7.曲线y=ex+x上点(0,1)处的切线方程为________。
正确答案:由曲线y=f(x)在其上点(x0,f(x0))的切线公式y-f(x0)=f’(x0)(x-x0),可知y-1=2(x-0),即所求切线方程为y=2x+1。
解析:注意点(0,1)在曲线y=ex+x上,又y’=ex+1,因此y’|x=0=2。
一、单项选择题(本大题共29小题,每小题2分,共58分)在每小题列出的四个备选项中只有一个是符合题目要求的,请用2B铅笔把答题卡上对应题目的答案字母按要求涂黑。
错选、多选或未选均无分。
1. 王老师为了提高全班同学的学习成绩,每周体育课上课前都会进行数学、语文、英语三个学科的单元测验,测验成绩达到80分以上的同学才有资格上体育课,单元测验成绩未达标的同学则需要在班级里自习,不能参加体育课。
李老师的做法()。
A.正确,帮助学生合理分配学习时间,提高学习效率B.错误,素质教育应面向全体学生,让学生全面发展C.正确,能够快速、有效地提高全班同学的学习成绩D.错误,应让全班同学体育课在班自习数、语、英三科2. 智育是要发展好奇心和理性思考的能力,而不是灌输知识;德育是要鼓励崇高的精神追求,而不是灌输规范;美育是要培育丰富的灵魂,而不是灌输技艺。
说明素质教育()A.要完全放手让学生自己探究和总结B.应改变学生以往的接受学习方式C.核心是智育、美育和德育的培养D.评价主体要做到多元化和综合化3. 下列关于“人的全面发展”的思想,表述错误的是()。
A.人的发展同其所处的社会生活条件没有联系B.旧式分工造成了人的片面发展C.机器大工业生产提供了人的全面发展的基础和可能D.教育与生产劳动相结合是培养全面发展的人的唯一途径4. 任老师在自己的教育学课堂上,针对不同基础的学生,通过不同的教学方法进行教学,使得每个学生都获得了发展。
任老师的做法体现了教师劳动的()。
A.长期性B.复杂性C.连续性D.创造性5. 小明在学校组织的一次社会实践活动中因个人问题出现了差错,造成任务的失败,班主任为了惩罚小明,让小明在上课时间打扫宿舍卫生。
班主任的做法侵犯了小明的()。
A.受教育权B.受完法定年限教育权C.人身权D.知识产权6. 小玥在上体育课时右臂骨折,经调查发现造成事故的主要原因是学校体育设施存在安全隐患,并没有及时处理。
按照《中华人民共和国教育法》规定,对学校主管人员应依法追究其()。
GRE(QUANTITATIVE)综合模拟试卷1(题后含答案及解析) 题型有:1. 2. 3.1.正确答案:A解析:In this question you are asked to compare the price at which Emma sold the bicycle with $140. From the information given, you can conclude that Emma spent a total of 75 + 27 = 102 dollars buying and repairing the bicycle and that she sold it for 40 percent more than the $102 she spent buying and repairing it. If you notice that 140 is 40 percent more than 100, you can conclude that 40 percent more than 102 is greater than 40 percent more than 100, and therefore, Quantity A is greater than Quantity B. The correct answer is Choice A.(If you solve the problem in this way, you do not have to calculate the value of Quantity A.)Another way to solve the problem is by explicitly calculating the value of Quantity A and comparing the result with $140 directly. Since 40 percent of 102 is(0.4)(102)= 40.8, it follows that Quantity A, the price at which Emma sold the bicycle, is 102.00 + 40.80 = 142.80 dollars. Thus Quantity A, $142.80, is greater than Quantity B, $140, and the correct answer is Choice A.2.正确答案:C解析:In this question you are asked to compare the area of the shaded region with 36. You are given that both PQRV and VRST are squares with sides of length 6. Therefore, you can conclude that the length of QS is 12, and the area of the shaded right triangle PQS is 1/2(12)(6), or 36. Thus Quantity A is equal to Quantity B, and the correct answer is Choice C.3.正确答案:A解析:Before making the comparison in this problem, you need to analyze the information given to see what it tells you about the value of Quantity A, the property tax in 2009 on a home in Town X that had an assessed value of $160,000. One way of doing this is to determine the value of the constant p and then use that value to calculate the tax on the home that had an assessed value of $160,000. Since it is given that a home that had an assessed value of $125,000 had a property tax of $2,500,you can conclude that p is equal to[*], or 2%. Once you know that the property tax is 2% of the assessed value, you can determine that tax on the home that had an assessed value of $160,000 was 2% of 160,000, or 3,200. The correct answer is Choice A. Another way to calculate the property tax on a home with an assessed value of $160,000 is by setting up a proportion. Because the tax rate is the same for each home in Town X, you can let the variable x represent the tax for the home assessed at $160,000 and solve for x as follows.[*]The correct answer is Choice A.4.正确答案:D解析:One way to approach this question is to plug in values for one of the variables and determine the corresponding value for the other variable. One way to plug in: Plug in easy values. For example, you can plug in x = 0 and find that the corresponding value of y is -1; then you can plug in y = 0 and find that the corresponding value of x is -1. Since in the first case x is greater than y and in the second case y is greater than x, the correct answer is Choice D, the relationship cannot be determined from the information given. A second way to plug in: If you prefer to always plug in values of x to determine corresponding values of y, you can begin by writing the equation x +y = -1 asy = -x - 1. Writing it in this form makes it easier to find the corresponding values of y. You can start by plugging in the value x = 0. For this value of x, the corresponding value of y is y = -1, and therefore, x is greater than y. If you continue plugging in a variety of values of x, some negative and some positive, you will see that sometimes x is greater than y and sometimes y is greater than x. If you inspect the equation y = —x - 1, you can conclude that since there is a negative sign in front of the x but not in front of the y, for each value of x that is greater than 0, the corresponding value of y is less than 0; therefore, for each x > 0, x is greater than y. What about negative values of x? A quick inspection of the equation y = —x - 1 allows you to conclude that if x < -1, then y > 0, so y is greater than x. So for some values of x and y that satisfy the equation, x is greater than y; and for other values, y is greater than x. Therefore, the relationship between the two quantities x and y cannot be determined from the information given, and the correct answer is Choice D.5.正确答案:B解析:You are given that three numbers, r, s, and t, are consecutive odd integers and that rIn the last step of the simplification, you can easily see that 3 In the last step of the simplification, you can easily see that 1 In the figure above, what is the value of?A.2B.3C.4D.5E.6正确答案:C解析:The sum of the measures, in degrees, of the three interior angles of any triangle is 180°. As shown in the figure, the three angles of the triangle have measures of x°, y°, and z°, so x + y + z = 180. Therefore,= 4, and the correct answer is Choice C.8.A certain store sells two types of pens: one type for $2 per pen and the other type for $3 per pen. If a customer can spend up to $25 to buy pens at the store and there is no sales tax, what is the greatest number of pens the customer can buy?A.9B.10C.11D.12E.20正确答案:D解析:It is fairly clear that the greatest number of pens that can be bought for $25 will consist mostly, if not entirely, of $2 pens. In fact, it is reasonable to begin by looking at how many of the $2 pens the customer can buy if the customer does not buy any $3 pens. It is easy to see that the customer could buy 12 of the $2 pens, with $1 left over. If the customer bought 11 of the $2 pens, there would be $3 left over with which to buy a $3 pen. In this case, the customer could still buy 12 pens. If the customer bought 10 of the $2 pens, there would be $5 left over. Only 1 of the $3 pens could be bought with the $5, so in this case, the customer could buy only 11 pens. As the number of $2 pens decreases, the total number of pens that the customer can buy with $25 decreases as well. Thus the greatest number of pens the customer can buy with $25 is 12. The correct answer is Choice D.9.If y = 3x and z = 2y, what is x + y + z in terms of x ?A.10xB.9xC.8xD.6xE.5x正确答案:A解析:It is not necessary to find the individual values of x, y, and z to answer the question. You are asked to rewrite the expression x + y + z as an equivalent expression in terms of x. This means that you need to use the information provided about y and z to express them in terms of the variable x. The variable y is already given in terms of x; that is, y = 3x; and because z = 2y, it follows that z =(2)(3x)= 6x. Using substitution, you can rewrite the expression as follows.x + y + z=x +(3x)+(6x)=(1 +3 + 6)x =10xThe correct answer is Choice A.10.A certain shipping service charges an insurance fee of $0.75 when shipping any package with contents worth $25.00 or less and an insurance fee of $1.00 when shipping any package with contents worth over $25.00. If Dan uses the shipping company to ship three packages with contents worth $18.25, $25.00, and $127.50, respectively, what is the total insurance fee that the company charges Dan to ship the three packages?A.1.75B.2.25C.2.5D.2.75E.3正确答案:C解析:Note that two of the packages being shipped have contents that are worth $25.00 or less. Therefore, each of them has an insurance fee of $0.75, for a total of $1.50. The third package has contents worth over $25.00, and it has an insurance fee of $1.00. Therefore, the total insurance fee for the three packages is $1.50 + $1.00 = $2.50, and the correct answer is Choice C.11.If 55 percent of the people who purchase a certain product are female, what is the ratio of the number of females who purchase the product to the number of males who purchase the product?A.11 to 9B.10 to 9C.9 to 10D.9 to 11E.5 to 9正确答案:A解析:Note that because 55 percent of the people who purchase the product are females, it follows that 45 percent of the people who purchase the product are males. Therefore, the ratio of the number of females who purchase the product to the number of males who purchase the product is 55 to 45, or 11 to 9, and the correct answer is Choice A.12.In the rectangular solid above, TU = 3, UV = 4, and VR = What is the area of the shaded rectangular region?______正确答案:10解析:To find the area of the shaded rectangular region, you need to multiply the length of the rectangular region by its width. In this question you are given the lengths of three edges: TU = 3, UV = 4, and VR = 2. Note that VR is the length of the shaded rectangle. To find the width of the shaded rectangle, you need to find either RS or VT. Note that VT lies on the front face of the rectangular solid. It is the hypotenuse of right triangle VUT. You know that UV = 4 and TU = 3, so by the Pythagorean theorem you can conclude that VT == 5. Therefore, the area of the shaded rectangularregion is(5)(2)= 10. The correct answer is 10.13.A list of numbers has a mean of 8 and a standard deviation of 2.If x is a number in the list that is 2 standard deviations above the mean, what is the value of x ? x=______正确答案:13解析:You are given that x is 2 standard deviations above the mean, 8. Because the standard deviation of the numbers in the list is 2.5, it follows that x is(2)(2.5), or 5 units above the mean 8. Therefore, x = 8 + 5 = 13, and the correct answer is 13.14.The circle graph above shows the distribution of 200,000 physicians by specialty. Which of the following sectors of the circle graph represent more than 40,000 physicians?Indicate all such sectors.A.PediatricsB.Internal MedicineC.SurgeryD.AnesthesiologyE.Psychiatry正确答案:A,B,C解析:One approach to solve this problem is to find out what percent of 200,000 is 40,000 and then compare this percent with the percents given in the circle graph. Because= 0.2, it follows that 40,000 is 20% of 200,000, and any specialty that has more than 20% of the distribution has more than 40,000 physicians. This is true for the specialties of pediatrics, internal medicine, and surgery. The correct answer consists of Choices A, B, and C.。
2023中考数学综合模拟习题一一.选择题(共12小题)1.下列图形中,既是轴对称图形又是中心对称图形的是()A.扇形B.正五边形C.菱形D.平行四边形2.下列计算正确的是()A.﹣a4b÷a2b=﹣a2b B.(a﹣b)2=a2﹣b2C.a2•a3=a6D.﹣3a2+2a2=﹣a23.下列实数中,最小的数是()A.B.0C.1D.4.下列说法正确的是()A.“明天降雨的概率为50%”,意味着明天一定有半天都在降雨B.了解全国快递包裹产生的包装垃圾数量适合采用全面调查(普查)方式C.掷一枚质地均匀的骰子,骰子停止转动后,6点朝上是必然事件D.一组数据的方差越大,则这组数据的波动也越大5.如图,直线a∥b,c,d是截线且交于点A,若∠1=60°,∠2=100°,则∠A=()A.40°B.50°C.60°D.70°5题图6题图6.已知一次函数y1=kx+b(k≠0)与反比例函数y2=(m>0)的图象如图所示,则当y1>y2时,自变量x满足的条件是()A.1<x<3B.1≤x≤3C.x>1D.x<37.受央视《朗读者》节目的启发的影响,某校七年级2班近期准备组织一次朗诵活动,语文老师调查了全班学生平均每天的阅读时间,统计结果如下表所示,则在本次调查中,全班学生平均每天阅读时间的中位数和众数分别是()每天阅读时间(小时)0.51 1.52人数89103A.2,1B.1,1.5C.1,2D.1,18.已知=3,则代数式的值是()A.B.C.D.9.已知圆内接正三角形的面积为,则该圆的内接正六边形的边心距是()A.2B.1C.D.10.如果关于x的不等式组的整数解仅有x=2、x=3,那么适合这个不等式组的整数a、b组成的有序数对(a,b)共有()A.3个B.4个C.5个D.6个11.如图,四边形AOEF是平行四边形,点B为OE的中点,延长FO至点C,使FO=3OC,连接AB、AC、BC,则在△ABC中S△ABO:S△AOC:S△BOC=()A.6:2:1B.3:2:1C.6:3:2D.4:3:211题图12题图12.如图,正方形ABCD的边长为2,P为CD的中点,连结AP,过点B作BE⊥AP于点E,延长CE交AD于点F,过点C作CH⊥BE于点G,交AB于点H,连接HF.下列结论正确的是()A.CE=B.EF=C.cos∠CEP=D.HF2=EF•CF二.填空题(共6小题)13.若2n(n≠0)是关于x的方程x2﹣2mx+2n=0的根,则m﹣n的值为.14.一个多边形的每一个外角都是18°,这个多边形的边数为.15.已知一组数据10,15,10,x,18,20的平均数为15,则这组数据的方差为.16.如图,在△ABC中,AF平分∠BAC,AC的垂直平分线交BC于点E,∠B=70°,∠F AE=19°,则∠C=度.16题图17题图17.如图,已知抛物线y=ax2﹣4x+c(a≠0)与反比例函数y=的图象相交于点B,且B点的横坐标为3,抛物线与y轴交于点C(0,6),A是抛物线y=ax2﹣4x+c的顶点,P点是x轴上一动点,当P A+PB最小时,P点的坐标为.18.已知函数y=使y=a成立的x的值恰好只有3个时,a的值为.三.解答题(共7题)19.(1)计算:﹣(1﹣)0+sin45°+()﹣1(2)先化简,再求值:÷(﹣),其中a=+2.20.某网络约车公司近期推出了”520专享”服务计划,即要求公司员工做到“5星级服务、2分钟响应、0客户投诉”,为进一步提升服务品质,公司监管部门决定了解“单次营运里程”的分布情况.老王收集了本公司的5000个“单次营运里程”数据,这些里程数据均不超过25(公里),他从中随机抽取了200个数据作为一个样本,整理、统计结果如下表,并绘制了不完整的频数分布直方图(如图).频数组别单次营运里程“x”(公里)第一组0<x≤572第二组5<x≤10a第三组10<x≤1526第四组15<x≤2024第五组20<x≤2530根据统计表、图提供的信息,解答下面的问题:(1)①表中a=;②样本中“单次营运里程”不超过15公里的频率为;③请把频数分布直方图补充完整;(2)请估计该公司这5000个“单次营运里程”超过20公里的次数;(3)为缓解城市交通压力,维护交通秩序,来自某市区的4名网约车司机(3男1女)成立了“交通秩序维护”志愿小分队,若从该小分队中任意抽取两名司机在某一路口维护交通秩序,请用列举法(画树状图或列表)求出恰好抽到“一男一女”的概率.21.如图所示,在平面直角坐标系中,一次函数y=kx+b(k≠0)与反比例函数y=(m ≠0)的图象交于第二、四象限A、B两点,过点A作AD⊥x轴于D,AD=4,sin∠AOD =,且点B的坐标为(n,﹣2).(1)求一次函数与反比例函效的解析式;(2)E是y轴上一点,且△AOE是等腰三角形,请直接写出所有符合条件的E点坐标.22.某销售商准备在某市采购一批丝绸,经调查,用10000元采购A 型丝绸的件数与用8000元采购B型丝绸的件数相等,一件A型丝绸进价比一件B型丝绸进价多100元.(1)求一件A型、B型丝绸的进价分别为多少元?(2)若销售商购进A型、B型丝绸共50件,其中A型的件数不大于B型的件数,且不少于16件,设购进A型丝绸m件.①求m的取值范围.②已知A型的售价是800元/件,销售成本为2n元/件;B型的售价为600元/件,销售成本为n元/件.如果50≤n≤150,求销售这批丝绸的最大利润w(元)与n(元)的函数关系式(每件销售利润=售价﹣进价﹣销售成本).23.如图,在直角三角形ABC中,∠ACB=90°,点H是△ABC的内心,AH的延长线和三角形ABC的外接圆O相交于点D,连结DB.(1)求证:DH=DB;(2)过点D作BC的平行线交AC、AB的延长线分别于点E、F,已知CE=1,圆O的直径为5.①求证:EF为圆O的切线;②求DF的长.24.如图,矩形ABCD中,AC=2AB,将矩形ABCD绕点A旋转得到矩形AB′C′D′,使点B的对应点B'落在AC上,B'C'交AD于点E,在B'C′上取点F,使B'F=AB.(1)求证:AE=C′E.(2)求∠FBB'的度数.(3)已知AB=2,求BF的长.25.如图,在等腰直角三角形ABC中,∠BAC=90°,点A在x轴上,点B在y轴上,点C(3,1),二次函数y=x2+bx﹣的图象经过点C.(1)求二次函数的解析式,并把解析式化成y=a(x﹣h)2+k的形式;(2)把△ABC沿x轴正方向平移,当点B落在抛物线上时,求△ABC扫过区域的面积;(3)在抛物线上是否存在异于点C的点P,使△ABP是以AB为直角边的等腰直角三角形?如果存在,请求出所有符合条件的点P的坐标;如果不存在,请说明理由.参考答案二.填空题第13题:12第14题:20 第15题:443第16题:24 第17题:(125,0)第18题:2三.解答题第19题:(1)原式=3√2 2(2)化简,可得,原式=a+2a−2,当a=+2时,原式=1+2√2第20题:(1)①48 ②0.73 ③(画图略)(2)750(3)12第21题:(1) 一次函数的解析式为y =−23x +2反比例函效的解析式为y =−12x(2)E 点坐标为(0,258)或(0,5)或(0,−5)第22题:(1) 解:设一件B 型丝绸的进价为x 元,则一件A 型丝绸的进价为(x+100)元,根据题意,可得,10000x +100=8000x解得,x =400经检验:x =400是原方程的解,且符合题意。
金融硕士(MF)金融学综合模拟试卷1(题后含答案及解析)全部题型 2. 不定项选择题3. 判断题4. 名词解释5. 计算题不定项选择题1.国际金币本位制的特点有:( )A.黄金是国际货币体系的基础。
B.黄金可以在各国间自由输入、输出,并被自由铸造成货币。
C.各国货币均可以按法定含金量与黄金自由兑换。
D.在该制度下,各国的国际收支调节具有对称性。
正确答案:A,B,C,D2.汇率决定的资产组合分析法将一国持有的资产分为:( )A.本国货币B.本国债券C.外币资产D.实物资产正确答案:A,B,C3.次贷危机的发生、蔓延原因有:( )A.银行相对于自有资本而言发放了过多的住房贷款。
B.美国的高利率使银行发放贷款的利息收入增加,从而刺激了住房抵押贷款的发放。
C.住房抵押贷款证券化后,又在市场上出售,从而使银行能获得更多的贷款资源。
D.金融创新助长了抵押贷款债券的发放和国际流通。
正确答案:A,C,D4.在国际收支的货币分析法中:( )A.假定一国的货币需求是稳定的。
B.一国的货币供给来自国内信贷和国际储备。
C.居民对手中货币余额的调节体现为国际收支。
D.当本国国内信贷扩张时,本国将出现国际收支盈余。
正确答案:A,B,C5.一国执行货币兑换管制的可能后果包括:( )A.能够节约外汇,用于集中进口本国必需品。
B.会形成规模较大的外汇黑市。
C.为获得外汇,本国厂商会倾向于高报出口、低报进口。
D.会形成经济租金,进一步引发不公平的再分配和寻租成本。
正确答案:A,B,D判断题6.根据多恩布什的“汇率超调论”,汇率之所以在受到货币冲击后会有过度反应,是因为资产价格的调整速度快于商品价格的调整速度。
( ) A.正确B.错误正确答案:A7.外债清偿率是指年还债(还本利息)总额与年商品和劳务出口收入之比,这个比例一般不应超过10%。
( )A.正确B.错误正确答案:B8.按P-e模型,如果没有技术进步和劳动生产率的提高,产出的增长会伴有物价水平的上升。
模拟卷一参考答案及解析一、单项选择题1.【答案】D。
【解析】“深山树木长不齐,荷花出水有高低”的意思是说,不可能所有学生都是一样的,表明学生与学生之间存在个体差异性。
2.【答案】B。
【解析】此题考察的是教师的教育理念,面对学生的学习疑问应该循循善诱,不应该讽刺学生,借此会打击学生的积极性,伤害学生的自尊心,故B为正确答案。
教育机智指的是教师面对突发情况采取的快速合理的措施,不符合此题情境,故A错;小雨对知识有疑惑不属于课堂无关行为,故C错;教师讽刺学生与教学任务的完成无关,故D错。
3.【答案】D。
【解析】此题考察对学生观的理解。
“以人为本”的学生观指出,学生是发展的人,作为教师应该用发展的眼光去看待学生,避免只关注学生的成绩,坚信每个学生都可以积极成长。
郭老师格外关照成绩好的学生,忽视了学生的发展性。
ABC题意未体现,故选D。
4.【答案】B。
【解析】此题考察个体具有差异性的知识。
素质教育强调促进学生的个性发展,个性是基于个人的特点和差异。
学生之间本身存在巨大的差异,而老师组织朗诵兴趣小组规定每个同学都必须参加,忽视了学生的意愿,也忽视了学生的个性差异。
故选B。
5.【答案】B。
【解析】《中华人民共和国教育法》第十五条规定,国务院教育行政部门主管全国教育工作,统筹规划、协调管理全国的教育事业。
6.【答案】C【解析】本题考查《中华人民共和国预防未成年人犯罪法》的相关规定。
依据《中华人民共和国预防未成年人犯罪法》第七章第五十条规定:“未成年人的父母或者其他监护人违反本法第十九条的规定,让不满十六周岁的未成年人脱离监护单独居住的,由公安机关对未成年人的父母或者其他监护人予以训诫,责令其立即改正。
”C项正确。
7.【答案】B【解析】本题考查的是学生的隐私权。
人身权是公民权利中最基本、最重要、内涵最为丰富的一项权利。
与学生密切相关的人身权包括生命健康权、隐私权、人身自由权和人格尊严权。
其中隐私权一般是指自然人享有的对自己的个人秘密和个人私生活进行支配并排除他人干涉的权利。
2023年中考化学综合训练(一)说 明:1.全卷满分70分,考试时间50分钟。
2.本试卷分为第I 卷(选择题)和第II 卷(非选择题)两部分,共四个大题,17个小题。
可能用到的相对原子质量:H —1 C —12 N —14 O —16 Fe —56 Cu —64 Ag —108 S —32第Ⅰ部分 (选择题 共 30 分)一、选择题(本题共 10 个小题,每小题 3 分,共 30 分。
每小题只有一个选项符合题意) 1.下列变化属于物理变化的是( )A .剩饭、剩菜变馊B .用稻草制作稻草人C .葡萄酿成酒D .鸡蛋壳上滴加食醋会冒气泡 2.保护环境,建设美丽广元。
下列说法正确的是( ) A .工业废水处理达标后再循环使用 B .田地秸秆就地焚烧后作农家肥使用 C .将化工厂烟囱加高以减少空气污染 D .将废弃塑料袋用土填埋以减少白色污染3.化学概念之间存在有并列、交叉、包含等不同关系。
下列选项符合M 、N 如图所示关系的是( )4.实验是学习化学的一条重要途径。
下列实验基本操作你认为正确的是( )A.氯化钠溶解 B .滴加液体 C .检验H 2纯度 D .收集CO 2 5.下列各组物质按单质、氧化物、混合物的顺序排列的是( ) A .锌粒、二氧化锰、矿泉水 B .石灰石、冰水混合物、液氧 C .水银、生铁、过氧化氢溶液 D .干冰、金刚石、白糖水A.化学与安全B.化学与健康(1)进入干涸的深井,一定要进行灯火实验(2)天然气泄漏,立即打开排风扇排气(1)喝牛奶有利于补钙(2)缺乏维生素C 会引起怀血病 C.化学与环境D.化学与材料(1)雨水呈酸性,所以叫酸雨(2)禁止使用农药和化肥,以减少水体污染(1)铝制易拉罐和废铁都属于金属材料 (2)合金、合成橡胶都属于有机合成材料482酸乙酯说法正确的是( )A .乙酸乙酯由4个碳原子、8个氢原子和2个氧原子构成B .乙酸乙酯中C 、H 、O 三种元素的质量比为4:8:2 C .乙酸乙酯的相对分子质量为88gD .乙酸乙酯中碳元素质量分数最大8.物质的鉴别、分离、提纯是化学重要的实验技能。
2022年中小学教师资格考试综合素质(中学)标准模拟卷(一)注意事项:1.考试时间为120分钟,满分为150分。
2.请按规定在答题卡上填涂、作答。
在试卷上作答无效,不予评分。
一、单项选择题(本大题共29小题,每小题2分,共58分)在每小题列出的四个备选项中只有一个是符合题目要求的,请用2B铅笔把答题卡上对应题目的答案字母按要求涂黑。
错选、多选或未选均无分1.不见用途,木材就无优劣之分。
红木是做家具的优材,但是不能用来盖房子;松木做家具是劣材,打桩可是优材。
正如斯皮尔曼教授一再强调的那样:“每一个人都可能是天才,每一个人都可能是白痴。
”该语句体现的是()。
A.学生是具有独立意义的主体B.学生是发展的人C.学生是具有个性与差异的人D.学生各阶段的发展要求不同2.某名牌师范大学毕业的小张回到家乡的一所普通中学任教。
小张根据课程标准认真上课,授课结束就认为该课程结束。
任教多年的老教师建议他对上课情况进行总结,查找自己上课的优缺点,小张一直)。
都置之不理。
从中可以看出小张(A.缺乏反思精神B.专业知识扎实C.缺乏合作精神D.缺乏创新精神3.孙老师总喜欢在考试成绩公布后给学生重新安排座位,他认为成绩好的学生就应该坐在前排或中间)。
的位置,成就差的学生就应该在后排或角落的位置。
孙老师的做法(A.正确,有很好的激励作用B.错误,成绩不是评价学生的唯一标准C.正确,成绩是评价学生的重要依据D.错误,成绩不能衡量学生的素质4.数学老师总以自己所授课程内容多,课时少为由,占用体育课、美术课等时间进行阶段测验、题目讲解等。
这种做法()。
A.正确,利于提高学生的数学成绩B.错误,不利于被占用老师的专业发展C.正确,利于提高班级整体教学质量D.错误,不利于学生的全面发展5.读初三的王玲成绩优异,一直是校一等奖学金的获得者。
今年王玲也榜上有名。
但是班主任却告诉她,她的奖学金将补贴班费。
因为班委会决定在中考后进行班级出游,但班费不足,经讨论班主任决定用她的奖学金补贴班费。
雅思写作(综合)模拟试卷1(题后含答案及解析) 题型有: 2. WRITING TASK 2WRITING TASK 2You should spend about 40 minutes on this task.1.In some countries young people are encouraged to work or travel for a year between finishing high school and starting university studies. Discuss the advantages and disadvantages for young people who decide to do this.正确答案:It is quite common these days for young people in many countries to have a break from studying after graduating from high school. The trend is not restricted to rich students who have the money to travel, but is also evident among poorer students who choose to work and become economically independent for a period of time. The reasons for this trend may involve the recognition that a young adult who passes directly from school to university is rather restricted in terms of general knowledge and experience of the world. By contrast, those who have spent some time earning a living or travelling to other places, have a broader view of life and better personal resources to draw on. They tend to be more independent, which is a very important factor in academic study and research, as well as giving them an advantage in terms of coping with the challenges of student life. However, there are certainly dangers in taking time off at that important age. Young adults may end up never returning to their studies or finding it difficult to readapt to an academic environment. They may think that it is better to continue in a particular job, or to do something completely different from a university course. But overall, I think this is less likely today, when academic qualifications are essential for getting a reasonable career. My view is that young people should be encouraged to broaden their horizons. That is the best way for them to get a clear perspective of what they are hoping to do with their lives and why. Students with such a perspective are usually the most effective and motivated ones and taking a year off may be the best way to gain this. 涉及知识点:写作2.Some people argue that teaching children of different abilities together benefits all of them. Others believe that intelligent children should be taught separately and given special treatment. Discuss both views and give your own opinion.正确答案:Today, it is quite common for many schools to practice an “elite students”policy, where the intelligent students are selected and given special treatment. Personally, I think this practice generally has more benefits on several different levels. The most obvious advantage lies in the fact that it allows teachers to employ different pedagogic strategies suited to the particular group of pupils. For example, for the more intelligent students who are more likely to have a quicker mindand stronger thirst for knowledge, teachers or tutors can include something more difficult or complex in their syllabus, catering for the special needs. On the other hand, for some other students who are not that intelligent and may be slow at learning or acquiring new knowledge or skills, a correspondingly suitable teaching strategy then is highly desirable. In this case, the separate teaching scheme is generally good for both groups of students, since teachers can thereafter design different courses and teaching methods catering for their special needs. Of course, those who are in support of integration teaching may argue that separating one from another is a violation of individual rights and that everyone should be treated equally regardless of their intelligence. I think the notion of equality is not the issue. The less intelligent students can be allowed to choose some advanced courses originally designed for those fast learners, and vice versa. Overall, I think it is not bad to separate students and nurture them individually. The authorities should take into consideration their pupils’ own preference as to being in or out of one particular group so that the benefits of this system can be maximised. 涉及知识点:写作3.Since science and technology are becoming more and more important in modern society, schools should spend more time on teaching these subjects rather than on arts and humanities. To what extent do you agree or disagree?正确答案:Some people claim that scientific and technological subjects should receive more emphasis in schools, given the importance of these subjects in the modern world. However, others think that this is not a sufficient reason to neglect education in arts and humanities. Obviously, there is little doubt that science and technology have had a great influence on modern society. The commonly cited examples of the Internet boom and the need to keep pace with a fast-changing global economy show us the importance of familiarising the next generation with science and technology. But does it mean that arts and humanities should take up a smaller proportion of a students’curriculum? Arts and humanities, it is argued, help students to gain a better vision of the past and how the world has, and continues to change over time. In other words, arts and humanities help them to think; to understand the mistakes made in the past, but also to reflect upon the great things humanity has accomplished. However, with the never-ending demand for new technologies and inventions, society is starting to pay less attention to the importance of arts and humanities. Therefore some people are wondering whether subjects in arts and humanities should give way to courses focusing on science and technology. It is imaginable that if this really happened, the younger generations today would lack the skills of critical thinking, knowledge of how the world turns around us, and suffer in their ability to read, reason and communicate. At a time when so much is happening to change the way we work and live, the way we relate to one another and the way we relate to the rest of the world, we actually need more knowledge and a better understanding of our shared history, philosophy and literature so that the lessons of the past are not lost. Therefore, I believe arts and humanities should not be given less attention in schools today, and the planning of a reasonable curriculumshould consider the overall benefits to the students. 涉及知识点:写作4.In many countries schools have severe problems with student behavior. What do you think are the causes of this? What solutions can you suggest?正确答案:Poor student behavior seems to be an increasingly widespread problem and I think that modern lifestyles are probably responsible for this. In many countries, the birth rate is decreasing sb that families are smaller with fewer children. These children are-often spoilt, not in terms of love and attention because working parents do not have the time for this, but in more material ways. They are allowed to have whatever they want, regardless of price, and to behave as they please. This means that the children grow up without consideration for others and without any understanding of where their standard of living comes from. When they get to school age they have not learnt any self control or discipline. They have less respect for their teachers and refuse to obey rules in the way that their parents did. Teachers continually complain about this problem and measures should be taken to combat the situation. But I think the solution to the problem lies with the families, who need to be more aware of the future consequences of spoiling their children. If they could raise them to be considerate of others and to be social, responsible individuals, the whole community would benefit. Perhaps parenting classes are needed to help them to do this, and high quality nursery schools could be established that would support families more in terms of raising the next generation. The government should fund this kind of parental support, because this is no longer a problem for individual families, but for society as a whole. 涉及知识点:写作。
线性代数综合练习题(一)
一、单项选择题
1. 对于n 阶可逆矩阵A ,B ,则下列等式中( A )不成立. (A) ()111
---⋅=B A AB (B) ())/1()/1(111
---⋅=B A AB (C) ()
1
1
1
---⋅=B
A
AB (D) ()
AB AB /11
=-
2. 若A 为n 阶矩阵,且03
=A ,则矩阵=--1)(A E ( B ).
(A )2
A A E +- (
B )2
A A E ++ (C )2
A A E -+ (D )2
A A E -- 3. 设A 是上(下)三角矩阵,那么A 可逆的充分必要条件是A 的主对角线元素为(C ). (A) 全都非负 (
B ) 不全为零 (
C )全不为零 (
D )没有限制 4. 设 3
3)(⨯=ij a A ,⎪⎪⎪⎭⎫
⎝⎛+++=133312
321131
1312
11
23
2221a a a a a
a a a a a a a B ,⎪⎪⎪⎭
⎫ ⎝⎛=1000010101P ,
⎪⎪⎪
⎭
⎫
⎝⎛=1010100012P ,那么( C ).
(A )B P AP =21 (B )B P AP =12 (C )B A P P =21 (D )B A P P =12 5. 若向量组m ααα,,,21 线性相关,则向量组内( A )可由向量组其余向量线性表示. (A )至少有一个向量 (B )没有一个向量 (C )至多有一个向量 (D )任何一个向量
6. 若⎪⎪⎪
⎭
⎫ ⎝⎛=210253143212A ,其秩=)(A R ( B ).
(A )1 (B )2 (C )3 (D )4
7. 若方程b AX =中,方程的个数小于未知量的个数,则有( B ). (A )b AX =必有无穷多解 (A )0=AX 必有非零解 (C )0=AX 仅有零解 (D )0=AX 一定无解 8. 若A 为正交阵,则下列矩阵中不是正交阵的是( B ). (A )1
-A (B )A 2 (C )4
A (D )T
A 9. 若满足条件( D ),则n 阶方阵A 与
B 相似.
(A )B A = (B ))()(B R A R = (C )A 与B 有相同特征多项式 (D )A 与B 有相同的特征值且n 个特征值各不相同 二、填空题
1. 若向量组
321,,ααα线性无关,则向量组321211,,αααααα+++是线性无
关 .
2. 设A 为4阶方阵,且3)(=A R ,*
A 是A 的伴随阵,则0=*
X A 的基础解系所含的解向量的个数是 3 .
3. 设A 为n 阶正交阵,且0>A ,则=A 3 .
4. 设()2,1,11-=α,()5,,22k =α, ()1,6,13-=α线性相关,则=k
3 .
5. 设⎪⎪⎪
⎭
⎫
⎝⎛=300050004A ,则=--1)2(E A
⎪⎪⎪
⎭
⎫ ⎝⎛10000003121
. 6. 设三阶方阵A 有特征值4,5,6,则=A 120 ,T
A 的特征值为
4.5.6 ,1
-A 的特征值为 6
1
5141,, . 三、计算题 1. 计算行列式
b a b b b b b a b b b b b a b b b b b a ----+----+a a
a a a a
b b b
b
a 000000-+=
40000000
00a a
a a
b b b a ==
2. 已知矩阵⎪⎪⎪⎭
⎫ ⎝⎛=200012021A ,求10
A .
首先求A的特征值
λ
λλ
λ---=
-20
0120
21E A =)1)(3)(2(λλλ+--- 1,3,2321-===λλλ
当21=λ时,由0)2(=-X E A 得,A 的对应于2的特征向量是⎪⎪⎪
⎭
⎫ ⎝⎛=1001ξ,
当32=λ时,有0)3(=-X E A 得,A 的对应于3的特征向量是⎪⎪⎪
⎭⎫ ⎝⎛=0112ξ,
当12-=λ时,有0)(=+X E A 得,A 的对应于1-的特征向量是⎪⎪⎪
⎭⎫ ⎝⎛-=0113ξ,
取⎪⎪⎪⎭
⎫ ⎝⎛=1001η⎪⎪⎪
⎭⎫ ⎝⎛-=⎪⎪⎪⎭⎫ ⎝⎛=01121,0112132ηη. 令()321,,ηηη=P ,则⎪⎪⎪⎭
⎫
⎝⎛-==-1321AP P AP P T
,所以
T
P P A 10
10132⎪⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎪⎭
⎫ ⎝
⎛+--+=1010
1101
1021
10
212000)13()
13(0)
13()13(
3. 设三阶方阵A 满足i i i A αα= )3,2,1(=i ,其中
T )2,2,1(1=α,T )1,2,2(2-=α,T )2,1,2(3--=α,求A .
3. 解:因为)3,2,1(==i i A i i αα,所以
⎪⎪⎪
⎭⎫ ⎝⎛=300020001),,(),,(321321ααααααA
因此1
3
21321),,(300020001),,(-⎪⎪⎪⎭
⎫ ⎝⎛=ααααααA 又),,(321ααα⎪⎪⎪⎭⎫ ⎝⎛---=212122221,所以1
321),,(-ααα⎪⎪⎪⎭⎫ ⎝⎛---=21212222191,
故=A ⎪⎪⎪⎭⎫ ⎝⎛---212122221⎪⎪⎪⎭⎫ ⎝⎛300020001⎪⎪⎪⎭⎫ ⎝⎛---21212222191⎪⎪⎪
⎭
⎫ ⎝⎛----=622250
207
31
4. λ取何值时,非齐次线性方程组
⎪⎩⎪
⎨⎧=-+=+-=-+1
610522321
321321x x x x x x x x x λλ (1) 有惟一解;(2)无解; (3)有无穷多解,并求其通解.
(1)解:)3)(5(6
10112
1
1
-+=---=λλλλD
(1)当0≠D ,即5-≠λ且3≠λ时,方程组有惟一解.
(2)当5-=λ时,⎪⎪⎪⎭⎫ ⎝⎛-----==1610155122151),(βA B −→−
r ⎪⎪⎪
⎭
⎫ ⎝⎛---100013902151 此时3)(,2)(==B R A R ,方程组无解,
(3)当3=λ时,⎪
⎪⎪⎭⎫ ⎝⎛---==1610153122131),(βA B −→−
r ⎪⎪⎪⎭
⎫
⎝
⎛--000010017
175
71778 此时2)()(==B R A R ,方程组有无限多个解.,并且通解为
⎪⎪⎪
⎭
⎫ ⎝⎛-+⎪⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎪⎭⎫ ⎝⎛10757871717321c x x x )(R c ∈. 四、证明题
1. 设A 为n 阶可逆阵,E A A =2
.证明A 的伴随阵A A =*
.\
1. 证:根据伴随矩阵的性质有
E A AA =*
又E A A =2
,所以2
A AA =*
,再由于A 可逆,便有A A =*
.
2. 若A ,B 都是n 阶非零矩阵,且0=AB .证明A 和B 都是不可逆的.
2. 证:假设A 可逆,即1-A 存在,以1
-A 左乘0=AB 的两边得0=B ,这与B 是n 阶非零
矩阵矛盾;类似的,若B 可逆,即1
-B 存在,以1
-B 右乘0=AB 的两边得0=A ,这与A
是n阶非零矩阵矛盾,因此,A和B都是不可逆的.。